1. The Mastery of Students on English Vocabulary Between Social Science Class and Natural Science Class of Eleveth Graders of SMAN 1 Kapuas Hilir a. The Description of the Data of Students in Social ScienceClass - A comparative study on english vocabulary

CHAPTER IV
RESULT OF THE STUDY

This chapter discussed the result of the studywhich consist of the data
finding and discussion.
A. Description of the Data
This section described the obtained data of the difference in the english
vocabulary mastery by Eleventh Graders of Social Science Class and Natural
Science Class at SMAN-1 Kapuas Hilir. The presented data consistof Mean,
Median, Modus, reliability value, and Standard Deviation.
1.

The Mastery of Students on English Vocabulary Between Social Science
Class and Natural Science Class of Eleveth Graders of SMAN 1 Kapuas
Hilir
a. The Description of the Data of Students in Social ScienceClass
The data presentation of the score of the students Social
Scienceshows in the table frequency distribution, the chart of frequency
distribution, the measurement of central tendency (mean, median, and mode)
and the measurement of deviation standard.
Table. 4.1 Description Data of Social Science Class

No
1
2
3
4

Name
D
JS
S
IR

Value
70
74
58
58

Range
B

B
D
D

5
6
7
No
8
9
10
11
12
13
14
15
16
17
18
19

20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

MR
A
JR
Name
RM
AR
NA
TBY
TKS
M
AR
YH
S
Z
YK
AP
CSD
YFA
WK

J
S
AP
MR
M
H
MH
EO
LD
A
T
DM
WE
M
IS
R
S

70
68

66
Value
70
66
82
82
72
74
68
70
74
78
78
82
72
62
68
88
86
76

76
54
74
82
66
74
82
84
80
72
76
70
68
58

B
C
C
Range
B

44
C
A
A
B
B
C
B
B
B
B
A
B
C
C
A
A
B
B
D

B
A
C
B
A
A
A
B
B
B
C
D

40
Total

DA

84
2912

A

Based on the data above, it can be seen that the students’ highest
score was 88 and the student’s lowest score was 54. To determine the range
of score, the class interval, and interval of temporary, the writer calculated
using formula as follows:
The Highest Score (H) = 88
The Lowest Score (L) = 54
The Range of Score (R) = H – L + 1
= 88 – 54 + 1
= 35
The Class Interval (K)

= 1 + (3.3) x Log n
= 1 + (3.3) x Log 40
= 1 + 4,293399
= 5,293399
=5
𝑅

Interval of Temporary (I)= =
𝐾

35
5

=7

Thus, the range of score was 35, the class interval was 5, and
interval of temporary was 7. It was presented using frequency distribution in
the following table:
Table 4.2 Frequency Distribution of Social Science Class Test Score
Class Interval Frequency
(K)
(I)
(F)

Mid
Point

Limitation
of Each

Frequency Frequency
Relative Cumulative

1
2
3
4
5

82– 88
75– 81
68– 74
61– 67
54– 60
Total

9
6
17
4
4
∑F = 40

(X)
85
78
71
64
57

Group
81,5 – 88,5
74,5 – 81,5
67,5 – 74,5
60,5 – 67,5
53,5 – 60,5

(%)
45
30
85
20
20
∑P = 200

(%)
45
75
160
180
200

The distribution of the score of students Social Science can also be
seen in the following Chart.
Figure 4.3The Frequency Distribution of the Score of Students
in Social Science Class

Frequency

The Frequency Distribution of the Score of Students Social
Science Class
18
16
14
12
10
8
6
4
2
0

17

9
6

81,5 - 88,5

74,5 - 81,5

67,5 - 74,5

4

4

60,5 - 67,5

53,5 - 60,5

Limitation of Each Group

According to the chart, the writer found that there are five kinds of
frequency in distribution score of students in social science class, there are
53,5 – 60,5. 60,5 – 67,5.67,5 – 74,5. 74,5 – 81,5. 81,5 – 88,5. There are 9
students who get the score 81,5 – 88,5, there are 6 students who get the

score in 74,5 – 81,5, there are 17 students who get the score in 67,5 – 74,5,
there are 4 students who get score in 60,5 – 67,5, and there are 4 students
who get the score in 53,5 – 60,5. It means that the frequency distribution of
the score of students in social class mostly occurred in 67,5 – 74,5.
The next step, the writer tabulated the scores into the table for the
calculation of mean, median and modus as follows :
Table 4.4 The Calculation of Mean, Median and Modus of
Students in Social Science Class Test Score
Mid
Point
(x)
85
78
71
64
57

Interval Frequency
(I)
(F)
82– 88
75– 81
68– 74
61– 67
54– 60

9
6
17
4
4
N=40

Fx

X’

FX’

Fka

Fkb

765
468
1207
256
228
∑Fx=2914

2
1
0
-1
-2

18
6
0
-4
-8
∑FX’= 12

9
15
32
36
40

40
31
25
8
4

1) Mean
Mx

=
=

∑fx
𝑁

2914
40

= 72,85
2) Median
1

Mdn

= ℓ +2

𝑁−𝑓𝑘𝑏
𝑓𝑖

= 67,5 +

𝑋𝑖

20−8
17

12

𝑋7

= 67,5 + 17 X 7

= 67,5+ 4,941
= 72,441
3) Modus
Mo

=u–

𝑓𝑏

𝑓𝑎 +𝑓𝑏

= 74,5 –

= 74,5 –

4

6+4
28
10

𝑥𝑖

𝑥7

= 74,5 – 2,8
= 71,7

4) Reliability
=

rxx
=
=

𝐾𝑠𝑥2 − 𝑋 (𝐾−𝑋)
𝑠𝑥2 (𝐾−1))

50 −68,9652 −73,1(50−73,1)
68,9652 (50−1)

237808 ,56125 +1,688,61
233052 ,390025

= 1,020

The calculation above shows thatthe mean value was 73,1, median
value was 72,441, modus value was 71,7, and reliabilty value was 1,020. So,
the test is reliable because rtest = 1,020 and rtable is 0,312. So rtest 1,020 > rtable
0,312 and the test is reliable.
Then, the writer tabulated the scores of student’s Social Science
Class into the table for the calculation of standard deviation as follows:
Table 4.5 The Calculation of Standard Deviation of the
Students in Social Science Class Test Score

Interval
(I)

Frequency
(F)

82– 88
75– 81
68– 74
61– 67
54– 60
Total

9
6
17
4
4
∑F = 40

Mid
Point
(X)
85
78
71
64
57

X’

Fx’

X’2

Fx’2

2
1
0
-1
-2

18
6
0
-4
-8
∑Fx’= 12

4
1
0
1
4

36
6
0
4
16
∑Fx’2= 62

Standard Deviation
SD = 𝑖

=7

∑𝑓𝑥 ′ 2
𝑁

62
40

–

-

∑(𝑓𝑥 )2

(12)2

𝑁

40

= 7 1,55 − 0,32

= 7 1,55 − 0,09

= 7 1,371

= 7 x 1,170897092
= 8,196
The tabulation above shows that the value of standard deviation is
8,196. Standard deviation used to measure the dissemination of the students

mean score, the result of the standard deviation is 8,196. It means that the
students mean score is dissemination.
b. The Description of the Data of Students in Natural Science Class
The data presentation of the score of the students Nature Science is
shows the table frequency distribution, the measurement of central tendency
(mean, median, and mode) and the measurement of deviation standard.In
order to analyze the vocabulary mastery by students Natural Science, it can
be first distributed by the following table:

Table 4.6 Description Data of Natural Science Class
No
1
2
3
4
5
6
7
8
9
10
11
12
13

Name
ARA
H
SR
TDS
MB
IK
OH
TO
Y
DLP
M
IPA
WA

Value
66
74
78
64
60
58
80
58
62
76
72
76
74

Range
C
B
B
C
C
D
A
D
C
B
B
B
B

14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
No
36
37
38
39
40
Total

CA
T
NS
AD
NSK
L
RG
SHN
A
WWA
Y
P
I
NH
S
N
W
MA
GAT
SJ
YA
W
Name
KA
JL
MP
AK
SA

78
60
68
74
76
50
54
66
64
62
68
64
56
80
56
60
84
80
82
50
70
70
Value
66
70
72
70
82
2730

B
C
C
B
B
D
D
C
C
C
C
C
D
A
D
C
A
A
A
D
B
B
Range
C
B
B
B
A

Based on the data above, it can be seen that the students’ highest
score is 84 and the student’s lowest score is 50. To determine the range of

score, the class interval, and interval of temporary, the writer calculated
using formula as follows:
The Highest Score (H)

= 84

The Lowest Score (L)

= 50

The Range of Score (R)

=H–L+1
= 84 –50 + 1
= 35

The Class Interval (K)

= 1 + (3.3) x Log n
= 1 + (3.3) x Log 40
= 1 + 4,293399
= 5,293399
=5
𝑅

= =

Interval of Temporary (I)

𝐾

35
5

=7
Thus, the range of score is 35, the class interval is 5, and interval of
temporary is 7. It is presented using frequency distribution in the following
table:
Table 4.7 Frequency Distribution of Natural Science Class Test
Score
Clas
s
(K)
1

Interval
(I)

Frequenc
y
(F)

Mid
Point
(X)

Limitation
of Each
Group

Frequency
Relative
(%)

78-84

8

81

77,5-84,5

40

Frequency
Cumulativ
e
(%)
40

2
3
4
5

71-77
64-70
57-63
50-56
Total

8
14
5
5
∑F = 40

74
67
60
53

70,5-77,5
63,5-70,5
56,5-63,5
59,5-56,5

40
70
25
25
∑P = 200

80
150
175
200

The distribution of the score of students Nature Science can also be
seen in the following Chart.
Figure 4.8The Frequency Distribution of the Score of Students
in Natura Science Class

Frequency

The Frequency Distribution of the Score of Students Natural
Science Class
16
14
12
10
8
6
4
2
0

14

8

8

77,5 - 84,5

70,5 - 77,5

63,5 - 70,5

5

5

56,5 - 63,5

59,5 - 56,5

Limitation of Each Group

According to the chart, the writer found that there are five kinds of
frequency in distribution score of students in social science class, there are
77,5-84,5. 70,5-77,5. 63,5-70,5. 56,5-63,5 and 59,5-56,5. There are 8
students who get the score 77,5-84,5, there are 8 students who get the score
in 70,5-77,5, there are 14 students who get the score in 63,5-70,5, there are
5 students who get score in 56,5-63,5, and there are 5 students who get the

score 59,5-56,5. It means that the frequency distribution of the score of
students in nature class mostly occurred in 63,5-70,5.
The next step, the writer tabulated the scores into the table for the
calculation of mean, median and modus as follows :
Table 4.9 The Calculation of Mean, Median and Modus of
Students in Natural Science Class
Interval
(I)

Frequency
(F)

78-84
71-77
64-70
57-63
50-56
Total

8
8
14
5
5
∑F = 40

Mid
Point
(X)
81
74
67
60
53

1) Mean
Mx

=
=

∑fx
𝑁

2731
40

= 68,27

2) Median
1

Mdn

= ℓ +2

𝑁−𝑓𝑘𝑏
𝑓𝑖

= 63,5 +

20−5
14

= 63,5 + 7,5
= 71

𝑋𝑖

𝑋7

Fx

X'

Fka

Fkb

648
592
938
300
265
∑Fx=2731

2
1
0
-1
-2

8
16
30
35
40

40
32
24
10
5

3) Modus
Mo

=u-

𝑓𝑏

𝑓𝑎 +𝑓𝑏

= 63,5 = 63,5 -

5

𝑥𝑖

8+5
35
13

𝑥7

𝑥7

= 63,5-18,84615384615385
= 44,654
4) Reliability
rxx

=

𝐾𝑠𝑥2 −𝑋 (𝐾−𝑋)

=

50 86,2982 −68,575 (50−68,575)

=

𝑠𝑥2 (𝐾−1))

86,2982 (50−1)

372367 ,2402 +1273 ,780625
364919 ,895396

= 1,024

The calculation above shows thatthe mean value was 68,5, median
value was 71, modus value was 44,65, and reliabilty value was 1,024. So,
the test is reliable because rtest = 1,024 and rtable is 0,312. So rtest 1,024 > rtable
0,312 and the test is reliable.
Then, the writer tabulated the scores of student’s Nature Science
Class into the table for the calculation of standard deviation as follows:
Table 4.1.1 The Calculation of the Standard Deviation of
Students in Natural Science Class
Mid
Interval Frequency
Point
(I)
(F)
(X)
78-84
8
81
71-77
8
74

X'

Fx'

X'2

Fx'2

2
1

16
8

4
1

32
8

64-70
57-63
50-56
Total

14
5
5
∑F = 40

67
60
53

0
-1
-2

0
-5
-10
∑Fx’= 9

0
1
4

0
5
20
∑Fx’2 = 65

Standard Deviation
SD = 𝑖

=7

∑𝑓𝑥 ′ 2
𝑁

65
40

-

-

(9)2

∑(𝑓𝑥 )2
𝑁

40

= 7x 1,625 − 0,2252

= 7x 1,625 − 0,050625
= 7 x 1,674375

= 7 x 1,2939764295
= 9,057
The tabulation above shows that the value of standard deviation is
9,057. Standard deviation used to measure the dissemination of the students
mean score, the result of the standard deviation is 9,057. It means that the
students mean score is dissemination.

1. The Differences and Similarities English Vocabulary Mastery Between
The Social Science Class and Natural Science Class Eleventh of Graders
at SMAN 1 Kapuas Hilir

a.

Description Data of Social Science Class Based on the Vocabulary
Mastery of Proper Noun, Verb, Adjective, Adverb
Table 4.1.2 Description Data of Social Science Class Based on the

Vocabulary Mastery of Proper Noun, Verb, Adjective, Adverb

No

Name

1
2
3
4
5
6
7

DEPRY
JEFFRY SANTOS
SELDY
INTAN RANTIAN
MUHAMMAD RIFKI
ANDIANTO
JULIADI RAHMADI
RIDWAN
MAULANA
ANDRI RAMADAN
NOR ADIANSYAH
TAWUN BUDI
YOKOB
TRY KUN. S
MARKANI
ADITYA ARISTO
YUSRAN HASANI
SUPIANI
ZAINAL
YAN KRISNA
ANTON PRAYOGO
CINDY SHINTYA
DITHA
YOHANES F.A
WAWAN
KRISTIANTO
JEFRYANTO
SURYADI
AGUNG P
M. RAFI'I
MACHDI

8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

Proper
Noun
100
89
84
94
84
94
100

Verb

Adjective

Adverb

63
72
72
63
45
54
45

77
88
33
11
66
55
77

18
36
18
27
72
45
18

100

63

77

18

89
100

54
63

88
88

18
63

94

81

100

45

100
94
89
94
94
94
94
84

81
72
54
72
54
63
54
63

77
100
77
66
77
100
77
100

9
18
36
27
54
45
72
81

84

63

55

72

94

63

55

9

94

72

77

9

100
100
100
100
68

72
54
63
90
72

88
77
88
66
22

81
100
36
27
36

28
29
30
31
32
33
34
35
36
37
38
39
40

HENGKY
MEGA HASANAH
EKA OCTAVIA
LEVIA DEVINIANTI
ALPIANOR
TRIWIRA
DONI MARTIN
WETIE ERLIANTI
MUHAJIRI
IGO SUSANTO
RIDUAN
SANTY
DESSY ADELINA
Total

100
100
94
100
100
100
94
100
100
100
100
84
89
3772

63
72
81
72
81
81
54
45
90
90
72
54
90
2682

77
88
55
77
100
77
77
66
77
66
44
33
88
2887

36
54
9
27
36
54
81
54
18
100
27
36
63
1685

Table 4.1.3Description Data of Social Science Class Based on the
Vocabulary Mastery of Proper Noun
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

Name
DEPRY
JEFFRY SANTOS
SELDY
INTAN RANTIAN
MUHAMMAD RIFKI
ANDIANTO
JULIADI RAHMADI
RIDWAN MAULANA
ANDRI RAMADAN
NOR ADIANSYAH
TAWUN BUDI
YOKOB
TRY KUN. S
MARKANI
ADITYA ARISTO
YUSRAN HASANI
SUPIANI
ZAINAL
YAN KRISNA

Proper Noun
100
89
84
94
84
94
100
100
89
100
94
100
94
89
94
94
94
94

19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

ANTON PRAYOGO
CINDY SHINTYA
DITHA
YOHANES F.A
WAWAN
KRISTIANTO
JEFRYANTO
SURYADI
AGUNG P
M. RAFI'I
MACHDI
HENGKY
MEGA HASANAH
EKA OCTAVIA
LEVIA DEVINIANTI
ALPIANOR
TRIWIRA
DONI MARTIN
WETIE ERLIANTI
MUHAJIRI
IGO SUSANTO
RIDUAN
SANTY
DESSY ADELINA
Total

84
84
94
94
100
100
100
100
68
100
100
94
100
100
100
94
100
100
100
100
84
89
3772

Based on the data above, it can be seen that the students’ highest
score of proper noun was 100 and the students’ lowest score of proper noun
was 68. To determine the range of score, the class interval, and interval of
temporary, the calculed using formula as follows:
The Highest Score (H)

=100

The Lowest Score (L)

= 68

The Range of Score (R)

=H–L+1

= 100 – 68 + 1
= 33
The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

33
7

=5
Thus, the range of score was 33, the class interval was 7, and interval
of temporary was 5. It was presented using frequency distribution in the
following table:
Table 4.1.4The Calculationof Mean of Social Science Class Based
on the Vocabulary Mastery of Proper Noun Score

No
1
2
3
4
5
6
7

Interval
(I)
98 – 102
93 – 97
88 – 92
83 – 87
78 – 82
73 – 77
68 – 72
Total

Mean
Mx

=

∑Fx
N

Frequency
(F)
18
12
4
5
0
0
1
∑F = 40

Mid point
(X)
100
95
90
85
80
75
70

Fx
1,800
1,140
360
425
0
0
70
∑Fx = 4,055

=

4,055
40

= 101,375
Table 4.1.5Description Data of Social Science Class Based on the
Vocabulary Mastery of Verb
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

Name
DEPRY
JEFFRY SANTOS
SELDY
INTAN RANTIAN
MUHAMMAD RIFKI
ANDIANTO
JULIADI RAHMADI
RIDWAN MAULANA
ANDRI RAMADAN
NOR ADIANSYAH
TAWUN BUDI YOKOB
TRY KUN. S
MARKANI
ADITYA ARISTO
YUSRAN HASANI
SUPIANI
ZAINAL
YAN KRISNA
ANTON PRAYOGO
CINDY SHINTYA
DITHA
YOHANES F.A
WAWAN KRISTIANTO
JEFRYANTO
SURYADI
AGUNG P
M. RAFI'I
MACHDI
HENGKY
MEGA HASANAH
EKA OCTAVIA

Verb
63
72
72
63
45
54
45
63
54
63
81
81
72
54
72
54
63
54
63
63
63
72
72
54
63
90
72
63
72
81

31
32
33
34
35
36
37
38
39
40

LEVIA DEVINIANTI
ALPIANOR
TRIWIRA
DONI MARTIN
WETIE ERLIANTI
MUHAJIRI
IGO SUSANTO
RIDUAN
SANTY
DESSY ADELINA
Total

72
81
81
54
45
90
90
72
54
90
2682

Based on the data above, it can be seen that the students’ highest
score of verb was 90 and the students’ lowest score of verb was 45. To
determine the range of score, the class interval, and interval of temporary, the
calculed using formula as follows:
The Highest Score (H)

= 90

The Lowest Score (L)

= 45

The Range of Score (R)

=H–L+1
= 90 – 45 + 1
= 46

The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

46
7

=7
Thus, the range of score was 46, the class interval was 7, and interval
of temporary was 7. It was presented using frequency distribution in the
following table:

Table 4.1.6The Calculation of Mean of Social Science Class
Based on the Vocabulary Mastery of Verb Score

No
1
2
3
4
5
6
7

Interval
(I)
87 – 93
80 – 86
73 – 79
66 – 72
59 – 65
52 – 58
45 – 51
Total

Frequency
(F)
4
5
0
10
10
8
3
∑F = 40

Mid point
(X)
90
83
76
69
62
55
48

Fx
360
415
0
690
620
440
144
∑Fx = 2,669

Mean
Mx

=
=

∑Fx
N

2,669
40

= 66,725
Table 4.1.7Description Data of Social Science Class Based on the
Vocabulary Mastery of Adjective
No
1

Name
DEPRY

Adjective
77

2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

JEFFRY SANTOS
SELDY
INTAN RANTIAN
MUHAMMAD RIFKI
ANDIANTO
JULIADI RAHMADI
RIDWAN MAULANA
ANDRI RAMADAN
NOR ADIANSYAH
TAWUN BUDI
YOKOB
TRY KUN. S
MARKANI
ADITYA ARISTO
YUSRAN HASANI
SUPIANI
ZAINAL
YAN KRISNA
ANTON PRAYOGO
CINDY SHINTYA
DITHA
YOHANES F.A
WAWAN
KRISTIANTO
JEFRYANTO
SURYADI
AGUNG P
M. RAFI'I
MACHDI
HENGKY
MEGA HASANAH
EKA OCTAVIA
LEVIA DEVINIANTI
ALPIANOR
TRIWIRA
DONI MARTIN
WETIE ERLIANTI
MUHAJIRI
IGO SUSANTO
RIDUAN
SANTY

88
33
11
66
55
77
77
88
88
100
77
100
77
66
77
100
77
100
55
55
77
88
77
88
66
22
77
88
55
77
100
77
77
66
77
66
44
33

40

DESSY ADELINA
Total

88
2887

Based on the data above, it can be seen that the students’ highest
score of adjective was 100 and the students’ lowest score of adjective was 11.
To determine the range of score, the class interval, and interval of temporary,
the calculed using formula as follows:
The Highest Score (H)

= 100

The Lowest Score (L)

= 11

The Range of Score (R)

=H–L+1
= 100 – 11 + 1
= 90

The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

90
7

= 13
Thus, the range of score was 90, the class interval was 7, and interval
of temporary was 13. It was presented using frequency distribution in the
following table:

Table 4.1.8The Calculation of Mean of Social Science Class
Based on the Vocabulary Mastery of Adjective Score

No
1
2
3
4
5
6
7

Interval
(I)
89 – 101
76 – 88
63 – 75
50 – 62
37 – 49
24 – 36
11 – 23
Total

Frequency
(F)
5
21
5
4
1
2
2
∑F = 40

Mid point
(X)
95
82
69
58
43
30
17

Fx
475
1,722
345
232
43
60
34
∑Fx = 2,911

Mean
Mx

=
=

∑Fx
N

2,911
40

= 72,775
Table 4.1.9Description Data of Social Science Class Based on the
Vocabulary Mastery of Adverb
No
1
2
3
4
5
6
7
8
9
10

Name
DEPRY
JEFFRY SANTOS
SELDY
INTAN RANTIAN
MUHAMMAD RIFKI
ANDIANTO
JULIADI RAHMADI
RIDWAN
MAULANA
ANDRI RAMADAN
NOR ADIANSYAH

Adverb
18
36
18
27
72
45
18
18
18
63

11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

TAWUN BUDI
YOKOB
TRY KUN. S
MARKANI
ADITYA ARISTO
YUSRAN HASANI
SUPIANI
ZAINAL
YAN KRISNA
ANTON PRAYOGO
CINDY SHINTYA
DITHA
YOHANES F.A
WAWAN
KRISTIANTO
JEFRYANTO
SURYADI
AGUNG P
M. RAFI'I
MACHDI
HENGKY
MEGA HASANAH
EKA OCTAVIA
LEVIA DEVINIANTI
ALPIANOR
TRIWIRA
DONI MARTIN
WETIE ERLIANTI
MUHAJIRI
IGO SUSANTO
RIDUAN
SANTY
DESSY ADELINA
Total

45
9
18
36
27
54
45
72
81
72
9
9
81
100
36
27
36
36
54
9
27
36
54
81
54
18
100
27
36
63
1685

Based on the data above, it can be seen that the students’ highest
score of adverb was 100 and the students’ lowest score of adverb was 9. To

determine the range of score, the class interval, and interval of temporary, the
calculed using formula as follows:
The Highest Score (H)

= 100

The Lowest Score (L)

=9

The Range of Score (R)

=H–L+1
= 100 – 9 + 1
= 92

The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

𝑅

=𝐾=

92
7

= 15

Thus, the range of score was 92, the class interval was 7, and interval
of temporary was 15. It was presented using frequency distribution in the
following table:
Table 4.2.1The Calculation of Mean of Social Science Class
Based on the Vocabulary Mastery of Adverb Score

No
1
2
3

Interval
(I)
99 – 203
84 – 98
69 – 83

Frequency
(F)
2
0
6

Mid point
(X)
151
91
76

Fx
302
0
456

54 – 68
39 – 53
24 – 38
9 – 23
Total

4
5
6
7

6
3
12
11
∑F = 40

61
46
31
16

366
138
372
176
∑Fx = 1,810

Mean
Mx

=
=

∑Fx
N

1,810
40

= 45,25
b. Description Data of Natural Science Class Based on the Vocabulary
Mastery of Proper Noun, Verb, Adjective, and Adverb
Table 4.2.2 Description Data of Natural Science Class Based on
the Vocabulary Mastery of Proper Noun, Verb, Adjective, and Adverb
No
1
2
3
4
5
6
7
8
9
10
11
12

Name
ANDRE RESTU
ADITIA
HANDOKO
SANDRO RIANTO
TRI DAUD
SAPUTRA
MONICA
BERLIANA
ISTI KOMAH
OKTAVI HARRY
TAMA OKTANIA
YULIANA
DESY LESTARI P.
MEGAWATI
INNES PUSPITA
APRILY

Proper
Noun

Verb

Adjective

Adverb

94

81

33

27

94
100

100
100

55
55

27
36

94

72

33

27

84

54

33

45

100
94
94
78
100
94

45
90
31
63
81
81

11
44
11
11
55
55

36
81
36
81
45
36

94

90

22

72

13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

WIDYA ASTUTI
CAHYA
ARIYANTI
TINTE
NOVITA SARI
ANA DELIMA
NINA SEPTIN
KRISTINA
LOLIANA
REBECKA
GLORIA
SUSAN HELDA
NATALIA
APRIANTO
WIRA WATI
ARISMA
YUNITA
PITALOKA
IDA
NURUL
HIDAYAH
SELLIN
NOVIA
WINARTI
MONA AUDIEA
GRESSELIN A.
TIARA
SITI JULEHA
YULIANI
ANGGRAINI
WINDA
KASPUL A.
JUNYALBY
LENGKEY
MEIDRIE
PRANATA
APRILIANO. K
SANTY AMIRA
Total

100

81

44

45

100

90

22

72

84
94
100

72
81
90

33
44
33

27
27
45

100

90

33

54

57

45

33

54

89

63

11

18

100

90

11

27

84

90

11

45

100

72

11

27

89
89
89

100
81
36

11
11
11

45
45
54

57

81

100

100

73
84
94
84

72
63
100
63

22
11
88
100

36
54
45
72

89

100

88

45

68

63

11

36

78

63

77

63

68
68

81
100

66
22

63
63

89

100

22

45

89

100

22

63

94
100
3530

90
100
3145

22
55
1443

36
54
1909

Table 4.2.3 Description Data of Natural Science Class Based on
the Vocabulary Mastery of Proper Noun

No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

Name
ANDRE RESTU
ADITIA
HANDOKO
SANDRO RIANTO
TRI DAUD
SAPUTRA
MONICA
BERLIANA
ISTI KOMAH
OKTAVI HARRY
TAMA OKTANIA
YULIANA
DESY LESTARI P.
MEGAWATI
INNES PUSPITA
APRILY
WIDYA ASTUTI
CAHYA
ARIYANTI
TINTE
NOVITA SARI
ANA DELIMA
NINA SEPTIN
KRISTINA
LOLIANA
REBECKA
GLORIA
SUSAN HELDA
NATALIA
APRIANTO
WIRA WATI
ARISMA
YUNITA
PITALOKA
IDA

Proper
Noun
94
94
100
94
84
100
94
94
78
100
94
94
100
100
84
94
100
100
57
89
100
84
100
89
89
89

27
28
29
30
31
32
33
34
35
36
37
38
39
40

NURUL
HIDAYAH
SELLIN
NOVIA
WINARTI
MONA AUDIEA
GRESSELIN A.
TIARA
SITI JULEHA
YULIANI
ANGGRAINI
WINDA
KASPUL A.
JUNYALBY
LENGKEY
MEIDRIE
PRANATA
APRILIANO. K
SANTY AMIRA
Total

57
73
84
94
84
89
68
78
68
68
89
89
94
100
3530

Based on the data above, it can be seen that the students’ highest
score of proper noun was 100 and the students’ lowest score of proper noun
was 57. To determine the range of score, the class interval, and interval of
temporary, the calculed using formula as follows:
The Highest Score (H)

=100

The Lowest Score (L)

= 57

The Range of Score (R)

=H–L+1
= 100 – 57 + 1
= 44

The Class Interval (K)

= 1 + (3,3) x Log n

= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7
Interval of Temporary (I)

=

𝑅

𝐾

=

44
7

=7
Thus, the range of score was 44, the class interval was 7, and interval
of temporary was 7. It was presented using frequency distribution in the
following table:
Table 4.2.4The Calculation of Mean of Natural Science Class
Based on the Vocabulary Mastery of Proper Noun Score

No
1
2
3
4
5
6
7

Interval
(I)
99 - 105
92 – 98
85 – 91
78 – 84
71 – 77
64 – 70
57 – 63
Total

Mean
Mx

=
=

∑Fx
N

3,548
40

= 88,7

Frequency
(F)
10
10
7
7
1
3
2
∑F = 40

Mid point
(X)
102
95
88
81
74
67
60

Fx
1,020
950
616
567
74
201
120
∑Fx = 3,548

Table 4.2.5Description Data of Natural Science Class Based on
the Vocabulary Mastery of Verb
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

Name
ANDRE RESTU
ADITIA
HANDOKO
SANDRO RIANTO
TRI DAUD
SAPUTRA
MONICA
BERLIANA
ISTI KOMAH
OKTAVI HARRY
TAMA OKTANIA
YULIANA
DESY LESTARI P.
MEGAWATI
INNES PUSPITA
APRILY
WIDYA ASTUTI
CAHYA
ARIYANTI
TINTE
NOVITA SARI
ANA DELIMA
NINA SEPTIN
KRISTINA
LOLIANA
REBECKA
GLORIA
SUSAN HELDA
NATALIA

Verb
81
100
100
72
54
45
90
31
63
81
81
90
81
90
72
81
90
90
45
63
90

22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

APRIANTO
WIRA WATI
ARISMA
YUNITA
PITALOKA
IDA
NURUL
HIDAYAH
SELLIN
NOVIA
WINARTI
MONA AUDIEA
GRESSELIN A.
TIARA
SITI JULEHA
YULIANI
ANGGRAINI
WINDA
KASPUL A.
JUNYALBY
LENGKEY
MEIDRIE
PRANATA
APRILIANO. K
SANTY AMIRA
Total

90
72
100
81
36
81
72
63
100
63
100
63
63
81
100
100
100
90
100
3145

Based on the data above, it can be seen that the students’ highest
score of verb was 100 and the students’ lowest score of verb was 31. To
determine the range of score, the class interval, and interval of temporary, the
calculed using formula as follows:
The Highest Score (H)

=100

The Lowest Score (L)

= 31

The Range of Score (R)

=H–L+1

= 100 – 31 + 1
= 70
The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

70
7

= 11
Thus, the range of score was 70, the class interval was 7, and interval
of temporary was 11. It was presented using frequency distribution in the
following table:
Table 4.2.6The Calculation of Mean of Natural Science Class
Based on the Vocabulary Mastery of Verb Score

No
1
2
3
4
5
6
7

Interval
(I)
97 - 107
86 – 96
75 – 85
64 – 74
53 – 63
42 – 52
31 – 41
Total

Mean
Mx

=

∑Fx
N

Frequency
(F)
9
8
8
4
7
2
2
∑F = 40

Mid point
(X)
102
91
80
69
58
47
36

Fx
918
728
640
276
406
94
72
∑Fx = 3,134

=

3,134
40

= 78,35
Table 4.2.7Description Data of Natural Science Class Based on
the Vocabulary Mastery of Adjective
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

Name
ANDRE RESTU
ADITIA
HANDOKO
SANDRO RIANTO
TRI DAUD
SAPUTRA
MONICA
BERLIANA
ISTI KOMAH
OKTAVI HARRY
TAMA OKTANIA
YULIANA
DESY LESTARI P.
MEGAWATI
INNES PUSPITA
APRILY
WIDYA ASTUTI
CAHYA
ARIYANTI
TINTE
NOVITA SARI
ANA DELIMA
NINA SEPTIN
KRISTINA
LOLIANA
REBECKA
GLORIA
SUSAN HELDA
NATALIA
APRIANTO
WIRA WATI

Adjective
33
55
55
33
33
11
44
11
11
55
55
22
44
22
33
44
33
33
33
11
11
11
11

24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

ARISMA
YUNITA
PITALOKA
IDA
NURUL
HIDAYAH
SELLIN
NOVIA
WINARTI
MONA AUDIEA
GRESSELIN A.
TIARA
SITI JULEHA
YULIANI
ANGGRAINI
WINDA
KASPUL A.
JUNYALBY
LENGKEY
MEIDRIE
PRANATA
APRILIANO. K
SANTY AMIRA
Total

11
11
11
100
22
11
88
100
88
11
77
66
22
22
22
22
55
1443

Based on the data above, it can be seen that the students’ highest
score of adjective was 100 and the students’ lowest score of adjective was 11.
To determine the range of score, the class interval, and interval of temporary,
the calculed using formula as follows:
The Highest Score (H)

= 100

The Lowest Score (L)

= 11

The Range of Score (R)

=H–L+1
= 100 – 11 + 1

= 90
The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

90
7

= 13
Thus, the range of score was 90, the class interval was 7, and interval
of temporary was 13. It was presented using frequency distribution in the
following table:

Table 4.2.8The Calculation of Mean of Natural Science Class
Based on the Vocabulary Mastery of Adjective Score

No
1
2
3
4
5
6
7

Interval
(I)
89 – 101
76 – 88
63 – 75
50 – 62
37 – 49
24 – 36
11 – 23
Total

Mean
Mx

=

∑Fx
N

Frequency
(F)
5
21
5
4
1
2
2
∑F = 40

Mid point
(X)
95
82
69
58
43
30
17

Fx
475
1,722
345
232
43
60
34
∑Fx = 2,911

=

2,911
40

= 72,775
Table 4.2.9Description Data of Natural Science Class Based on
the Vocabulary Mastery of Adverb
No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

Name
ANDRE RESTU
ADITIA
HANDOKO
SANDRO RIANTO
TRI DAUD
SAPUTRA
MONICA
BERLIANA
ISTI KOMAH
OKTAVI HARRY
TAMA OKTANIA
YULIANA
DESY LESTARI P.
MEGAWATI
INNES PUSPITA
APRILY
WIDYA ASTUTI
CAHYA
ARIYANTI
TINTE
NOVITA SARI
ANA DELIMA
NINA SEPTIN
KRISTINA
LOLIANA
REBECKA
GLORIA
SUSAN HELDA
NATALIA
APRIANTO
WIRA WATI

Adverb
27
27
36
27
45
36
81
36
81
45
36
72
45
72
27
27
45
54
54
18
27
45
27

24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

ARISMA
YUNITA
PITALOKA
IDA
NURUL
HIDAYAH
SELLIN
NOVIA
WINARTI
MONA AUDIEA
GRESSELIN A.
TIARA
SITI JULEHA
YULIANI
ANGGRAINI
WINDA
KASPUL A.
JUNYALBY
LENGKEY
MEIDRIE
PRANATA
APRILIANO. K
SANTY AMIRA
Total

45
45
54
100
36
54
45
72
45
36
63
63
63
45
63
36
54
1909

Based on the data above, it can be seen that the students’ highest
score of adverb was 100 and the students’ lowest score of adverb was 18. To
determine the range of score, the class interval, and interval of temporary, the
calculed using formula as follows:
The Highest Score (H)

= 100

The Lowest Score (L)

= 18

The Range of Score (R)

=H–L+1
= 100 – 18 + 1

= 83
The Class Interval (K)

= 1 + (3,3) x Log n
= 1 + (3,3) x Log 40
= 1 + 5,9020599913
= 6,9020599913
=7

Interval of Temporary (I)

=

𝑅

𝐾

=

83
7

= 13
Thus, the range of score was 83, the class interval was 7, and interval
of temporary was 13. It was presented using frequency distribution in the
following table:

Table 4.3.1The Calculation of Mean of Natural Science Class
Based on the Vocabulary Mastery of Adverb Score
No
1
2
3
4
5
6
7

Interval
(I)
96 – 108
83 – 95
70 – 82
57 – 69
44 – 56
31 – 43
18 – 30

Frequency
(F)
1
0
5
4
15
7
8

Mid point
(X)
102
89
76
63
50
37
24

Fx
102
0
380
252
750
259
192

Total

∑F = 40

∑Fx = 1,935

Mean
Mx

=
=

∑Fx
N

1,935
40

= 48,375
Based on the calculation above, the mean score of Social Scince
Class and Natural Science Class according to the types of test is the mean of
proper noun for social science class was 101,3 and natural science class was
89. So, the social science class got higher score than natural science class in
proper noun test. Then for the mean verb of social science class was 67 and
natural science class was 78,3. So, the natural science class was higher than
social science class in verb test. Next, the mean score of social science class
in adjective was 73 and natural science class 73. So, the social science class
and natural science class get similar score. The last the mean score of adverb
for social science class was 45,2 and natural science class was 48,3. So, the
naural science class higher than social science class.
B. Test of the Statistical Analysis
Meanwhile, the calculation of Ttest using SPSS 17.0 Program can be
seen in the following table :
Group Statistics
Group
Social Science Class

N

Mean

40

72.80

Std.
Deviation
8.200

Std. Error
Mean
1.297

Group Statistics
Group
Social Science Class
Natural Science Class

N

Mean

40
40

72.80
68.25

Std.
Deviation
8.200
9.173

Std. Error
Mean
1.297
1.450

Independent Samples Test
Nilai Ujian

Levene's Test
for Equality of
Variances
t-test for
Equality of
Means

F
Sig.
T
Df
Sig. (2-tailed)
Mean Difference
Std. Error Difference
95% Confidence Lower
Interval of the
Upper
Difference

Equal variances
assumed
1.101
.297
2.339
78
.022
4.550
1.945
.677
8.423

Equal
variances
not
assumed

2.339
77.040
.022
4.550
1.945
.676
8.424

The result of t test using SPSS 17.0 supported the interpretation of t-test
result from manual calculation. It shows from the table above that the tobservedwas
2,339. It was also higher than ttable at 5% (1,991) level of significance. Therefore,
it could be interpreted that Ho stating that there is no significant difference in
English vocabulary mastery between eleventh graders of social science class and
natural science class was rejected and Ha stating that there is

a significant

difference in English vocabulary mastery between eleventh graders of social
science class and natural science class at SMAN-1 Kapuas Hilir was accepted at
5% level of significance.
C. Result of the Data Analysis
In order tocalculate the ttest, the writer used both manual calculation.
Both results are expected to support the correct calculation each other.
After knowing Standard Deviation of group I and group II, the writer
calculated the “to” value to examine the hypothesis. But, first of all the writer
calculated the variance homogeneity in order to adjust the formula in calculating
the “to” value, becausesome formula used to examine the comparative hypothesis
with two sample,thereis Fisher formula.Furthermore, in order to ease the
calculation of test of variance homogeneity and test of hypothesis, the writer
makes a table to compare the N (number of sample), mean, variance, and
deviation standard of two groups.
Table. 4.3.2 The Data of Test Scores of Eleventh Graders of Social
Science Class and Natural Science Class at SMAN-1 Kapuas Hilir

No
1
2
3
4

The students’ score in social
science class
70
74
58
58

The students’ score in natural
science class
66
74
78
64

5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

70
68
66
70
66
82
82
72
74
68
70
74
78
78
82
72
62
68
88
86
76
76
54
74
82
66
74
82
84
80
72
76
70
68
58
84

60
58
80
58
62
76
72
76
74
78
60
68
74
76
50
54
66
64
62
68
64
56
80
56
60
84
80
82
50
70
70
66
70
72
70
82

N
Mx
S1
S12

40
72,85
8,196
68,965

40
68,27
9,057
86,298

1. Variance Homogeneity
F=
=

𝑇𝑕𝑒 𝐵𝑖𝑔𝑔𝑒𝑠𝑡 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒

𝑇𝑕𝑒 𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒

86,298

68,965

= 1,251
Moreover, the result variance homogeneity was compared with Ftable on numerator df ( 40-1 = 39) and denominator df (40-1= 39). Based on
those df with significant 5%, than the percentage of F table was 1,75. It found
that Fvalue was smaller than Ftable (1,251< 1,75). Therefore, it can be said that
the variance of those two groups was homogeneous.
Since the number of sample of those two groups was same ( N1 =
N2 ), and the variance was homogen. Thus, the testing of t observed was used
Fisherformula.
2. Testing of Normality test
Normality test is a test to know about what the writing test had given
to the students normally, it shows about:
1) Normality test of Students in Social Science Class
Table 4.3.3 Normality test of Students in Social Science Class

No

X

Z

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

54
58
58
58
62
66
66
66
68
68
68
68
70
70
70
70
70
72
72
72
74
74
74
74
74
76
76
76
78
78
80
82
82
82
82
82

-2,2227
-1,7498
-1,7498
-1,7498
-1,2768
-0,8039
-0,8039
-0,8039
-0,5675
-0,5675
-0,5675
-0,5675
-0,3310
-0,3310
-0,3310
-0,3310
-0,3310
-0,0945
-0,0945
-0,0945
0,1418
0,1418
0,1418
0,1418
0,1418
0,3783
0,3783
0,3783
0,6148
0,6148
0,8512
1,0877
1,0877
1,0877
1,0877
1,0877

Table
Z
0,5040
0,5000
0,5000
0,5000
0,4168
0,3058
0,3058
0,3058
0,2119
0,2119
0,2119
0,2119
0,1151
0,1151
0,1151
0,1151
0,1151
0,0158
0,0158
0,0158
0,1357
0,1357
0,1357
0,1357
0,1357
0,2420
0,2420
0,2420
0,3446
0,3446
0,3594
0,4440
0,4440
0,4440
0,4440
0,4440

F(Zi)

F(kum)

S(Zi)

0,0122
0,0401
0.0401
0,0401
0,1056
0,1977
0,1977
0,1977
0,2912
0,2912
0,2912
0,2912
0,3632
0,3632
0,3632
0,3632
0,3632
0,4801
0,4801
0,4801
0,5596
0,5596
0,5596
0,5596
0,5596
0,6368
0,6368
0,6368
0,7422
0,7422
0,8023
0,8531
0,8531
0,8531
0,8531
0,8531

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

0,025
0,05
0,075
0,1
0,125
0,15
0,175
0,2
0,225
0,25
0,275
0,3
0,325
0,35
0,375
0,4
0,425
0,45
0,475
0,5
0,525
0,55
0,575
0,6
0,625
0,65
0,675
0,7
0,725
0,75
0,775
0,8
0,825
0,85
0,875
0,9

F(zi)S(zi)
-0,0128
-0,0099
-0,0349
-0,0599
-0,0194
0,0477
0,0227
-0,0023
0,0662
0,0412
0,0162
-0,0088
0,0382
0,0132
-0,0118
-0,0368
-0,0618
0,0301
0,0051
-0,0199
0,0346
0,0096
-0,0154
-0,0404
-0,0654
-0,0132
-0,0382
-0,0632
0,0172
-0,0078
0,0273
0,0531
0,0281
0,0031
-0,0219
-0,0469

37
38
39
40
Total
Mean
STDEV
Ltest
Ltable

84
84
86
88
2912
72,80
8,196
0,0662
0,139

1,3241
1,3241
1,5606
1,7971

0,4562
0,4562
0,4522
0,4602

0,9115
0,9115
0,9394
0,9599

37
38
39
40

0,925
0,95
0,975
1

-0,0135
-0,0385
-0,0356
-0,0401

The table shows that Ltest= 0,0662< Ltable= 0,139, then the data of
students social science class distributed normally.
2) Normality test of Studentsin Natural Science Class
Table 4.3.4 Normality test of Students in Natural Science Class
No

X

Z

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

50
50
54
56
56
58
58
60
60
60
62
62
64
64
64
66
66
66

-2,0778
-2,0778
-1,6224
-1,3947
-1,3947
-1,1670
-1,1670
-0,9393
-0,9393
-0,9393
-0,7116
-0,7116
-0,4838
-0,4838
-0,4838
-0,2561
-0,2561
-0,2561

Table
Z
0,4325
0,4325
0,4207
0,4129
0,4129
0,4052
0,4052
0,3409
0,3409
0,3409
0,2327
0,2327
0,1131
0,1131
0,1131
0,0228
0,0228
0,0228

F(Zi)

F(kum)

S(Zi)

0,0202
0,0202
0,0495
0,0885
0,0885
0,1251
0,1251
0,1711
0,1711
0,1711
0,2266
0,2266
0,3264
0,3264
0,3264
0,4013
0,4013
0,4013

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

0,025
0,05
0,075
0,1
0,125
0,15
0,175
0,2
0,225
0,25
0,275
0,3
0,325
0,35
0,375
0,4
0,425
0,45

F(zi)S(zi)
-0,0048
-0,0298
-0,0255
-0,0115
-0,0365
-0,0249
-0,0499
-0,0289
-0,0539
-0,0789
-0,0484
-0,0734
0,0014
-0,0236
-0,0486
0,0013
-0,0237
-0,0487

19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Total
Mean
STDEV
Ltest
Ltable

68
68
70
70
70
70
72
72
74
74
74
76
76
76
78
78
80
80
80
82
82
84
2730
68,25
9,057
0,0672
0,139

-0,0284
-0,0284
0,1992
0,1992
0,1992
0,1992
0,4269
0,4269
0,6546
0,6546
0,6546
0,8823
0,8823
0,8823
1,1100
1,1100
1,3378
1,3378
1,3378
1,5655
1,5655
1,7932

0,0179
0,0179
0,1093
0,1093
0,1093
0,1093
0,2033
0,2033
0,3015
0,3015
0,3015
0,3121
0,3121
0,3121
0,3228
0,3228
0,4052
0,4052
0,4052
0,4190
0,4190
0,4247

0,4801
0,4801
0,5596
0,5596
0,5596
0,5596
0,6736
0,6736
0,7422
0,7422
0,7422
0,8023
0,8023
0,8023
0,8749
0,8749
0,9115
0,9115
0,9115
0,9394
0,9394
0,9599

19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

0,475
0,5
0,525
0,55
0,575
0,6
0,625
0,65
0,675
0,7
0,725
0,75
0,775
0,8
0,825
0,85
0,875
0,9
0,925
0,95
0,975
1

0,0051
-0,0199
0,0346
0,0346
-0,0154
-0,0404
0,0486
0,0236
0,0672
0,0422
0,0172
0,0523
0,0273
0,0023
0,0499
0,0249
0,0365
0,0115
-0,0135
-0,0106
-0,0356
-0,0401

The table shows that Ltest= 0,0672 < Ltable= 0,139, then the data of
students nature science class distributed normally.
3. Testing of tobserved (to)

Table. 4.3.5 The Data of Test Scores of Students Eleventh
Graders of Social Science Class and Natural Science Class at SMAN-1
Kapuas Hilir
Scores
X1
70
74
58
58
70
68
66
70
66
82
82
72
74
68
70
74
78
78
82
72
62
68
88
86
76
76
54
74
82
66
74
82
84

X2
66
74
78
64
60
58
80
58
62
76
72
76
74
78
60
68
74
76
50
64
66
64
62
68
64
56
80
56
60
84
80
82
50

X1
-2,8
+1,2
-14,8
-14,8
-2,8
-4,8
-6,8
-2,8
-6,8
+9,2
+9,2
-0,8
+1,2
-4,8
-2,8
+1,2
+5,2
+5,2
+9,2
-0,8
-10,8
-4.8
15.2
13,2
3,2
3,2
-18,8
+1,2
+9,2
-6,8
+1,2
+9,2
+11,2

X2
-2,25
+5,75
+9,75
-4,25
-8,25
-10,25
+11,75
-10,25
-6,25
+7,75
+3,75
+7,75
+5,75
+9,75
-8,25
-0,25
+5,75
+7,75
-18,25
-14,25
-2,25
-4,25
-6,25
-0,25
-4,25
-12,25
+11,75
-12,25
-8,25
+15,75
+11,75
+13,75
-18,25

X12
7,84
1,44
219,04
219,04
7,84
23,04
46,24
7,84
46,24
84,64
84,64
0,64
1,44
23,04
7,84
1,44
27,04
27,04
84,64
0,64
116,64
23,04
231,04
174,24
10,24
10,24
353,44
1,44
84,64
46,24
1,44
84,64
125,44

X22
5,0625
33,0625
95,0625
18,0625
68,0625
105,0625
138,0625
105,0625
39,0625
60,0625
14,0625
60,0625
33,0625
95,0625
68,0625
0,0625
33,0625
60,0625
333,0625
203,0625
5,0625
18,0625
39,0625
0,0625
18,0625
150,0625
138,0625
150,0625
68,0625
248,0625
138,0625
189,0625
333,0625

80
72
76
70
68
58
84
∑X= 2912

t0 =

t=

t=

t=
t=
t=

70
70
66
70
72
70
82
∑Y= 2730

𝐌𝟏 − 𝐌𝟐

∑ 𝐗 𝟐 + ∑𝐗 𝟐
𝟏
𝟐
𝐍𝟏 + 𝐍𝟐 −𝟐

+7,2
-0,8
+3,2
-2,8
-4,8
-14,8
+11,2
∑X1=
249E,14

𝐍𝟏 + 𝐍𝟐
𝐍𝟏 . 𝐍𝟐

72,8− 68,25
2622 ,4 +3281 ,5
40 + 40 −2

40 + 40
40 . 40

4,55
5903 ,9
78

X

80
1600

4,55
75,691 X 0,05
4,55
3,78455
4,55
1,945

t = 2,339
The degree of Freedom
Df = N1 + N2 – 2
= 40 + 40 – 2
= 78
Df 78 at 5% level of significant = 1,991
T0 = 2,339> Ttable = 1,991

+1,75
+1,75
-2,25
+1,75
+3,75
+1,75
+13,75
∑X2= 0

51,58
0,64
10,24
7,84
23,04
219,04
125,44
∑X12=
2622,4

3,0635
3,0625
5,0625
3,0625
14,0625
3,0625
189,0625
∑X22=
3281,5

(Ha was accepted)
Based on the result above, it can be presented by the following table:
Table 4.3.6 The Result of Tobserved
t0

tt

Df

2,339

1,991

78

Where :
to : The value of tobserved
tt : The value of ttable
Since the calculated value of tobserved (2,339) was higher than ttable at 5%
(1,991) significant level or 2,339>1,991, it could be interpreted that there is no
significant difference in English vocabulary mastery between eleventh graders of
social science class and natural science class, thus Ho (Null Hypotesis) was
rejected and there isa significant difference in English vocabulary mastery
between eleventh graders of social science class and natural science class, thus the
Ha (Alternative hypotesis) was accepted. It means that there is a significant
difference in English vocabulary mastery between eleventh graders of social
science class and natural science class at SMAN-1 Kapuas Hilir.